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Question: How do you solve \(2{\sin ^2}x = 1\)?...

How do you solve 2sin⁡2x=12{\sin ^2}x = 1?

Explanation

Solution

This is a trigonometric problem and it can be solved with basic properties of trigonometry. Here, we have to find the value of x. For which we need to simplify the equation and bring it to the form of sin x. And then solve it by placing the sin value.

Complete step by step solution:
The given equation is 2sin⁡2x=12{\sin ^2}x = 1
We will take the co-efficient of sin⁡2x{\sin ^2}x i.e., 22 on the other side of equal to
∴sin⁡2x=12\therefore {\sin ^2}x = \dfrac{1}{2}
Now, we will take the square root of both the side
⇒sin⁡x=±12\Rightarrow \sin x = \pm \dfrac{1}{{\sqrt 2 }}
Here first we consider sin⁡x=12\sin x = \dfrac{1}{{\sqrt 2 }}
Now put sin⁡45∘=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}, we get,
⇒sin⁡x=sin⁡π4\Rightarrow \sin x = \sin \dfrac{\pi }{4}
Since 45∘=π4{45^ \circ } = \dfrac{\pi }{4}in radian.
Similarly, now considersin⁡x=−12\sin x = - \dfrac{1}{{\sqrt 2 }}
Now put sin⁡45∘=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}, but as it is minus it will be sin⁡(−π4)\sin ( - \dfrac{\pi }{4}), we get,
⇒sin⁡x=sin⁡(−π4)\Rightarrow \sin x = \sin ( - \dfrac{\pi }{4})
When we solve the above equation
⇒sin⁡x=sin⁡(π−π4)\Rightarrow \sin x = \sin (\pi - \dfrac{\pi }{4})
⇒sin⁡x=sin⁡(4π−π4)\Rightarrow \sin x = \sin (\dfrac{{4\pi - \pi }}{4})
So, after simplifying we get,
⇒sin⁡x=sin⁡3π4\Rightarrow \sin x = \sin \dfrac{{3\pi }}{4}
So we will consider all the values in all the quadrants of sin⁡x\sin x where x=π4,3π4,5π4,7π4,.............x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4},.............
∴sin⁡x=sin⁡(π4+π2n)\therefore \sin x = \sin (\dfrac{\pi }{4} + \dfrac{\pi }{2}n)
where, n is 1,2,3,4,…… or we can say n∈Rn \in R
∴x=π4+π2n\therefore x = \dfrac{\pi }{4} + \dfrac{\pi }{2}n.

Additional Information:
The sin⁡x\sin x function gives progressive values that mean from sin⁡0∘\sin {0^ \circ } to sin⁡90∘\sin {90^ \circ } the value of sin⁡x\sin x will increase. So, some basic values of sin⁡θ\sin \theta where θ\theta is 0∘,30∘,45∘,60∘{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }and 90∘{90^ \circ }.
∴sin0∘=0,sin⁡30∘=12,sin⁡45∘=12,sin⁡60∘=32,sin⁡90∘=1\therefore sin{0^ \circ } = 0,\sin {30^ \circ } = \dfrac{1}{2},\sin {45^ \circ } = \sqrt {\dfrac{1}{2}} ,\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {90^ \circ } = 1.

Note: In order to solve this problem, we must learn about the values of sin⁡θ\sin \theta where θ\theta can be any degree. We also have to know about the values of sin⁡30∘,sin⁡45∘\sin {30^ \circ },\sin {45^ \circ } and sin⁡90∘\sin {90^ \circ }. The trigonometric function sin⁡x\sin x range lies [−1,1][ - 1,1] and it is a closed interval that means it includes -1 as well as 1. All the real numbers are the domain of the trigonometric function sin⁡x\sin x .
The sign convention should be taken care of because the value of sin x has a square root in it. And we already know that the square of a positive or negative number is always positive.
We should always keep in mind that the square root has plus/minus signs which state that we require both the positive integer as well as negative integers.